Problem 74 - Calculus Early Transcendentals

Question

In polar coordinates an equation of an ellipse with eccentricity \(0 < e < 1\) and semimajor axis \(a\) is \(r=\frac{a\left(1-e^{2}\right)}{1+e \cos \theta}\) a. Write the integral that gives the area of the ellipse. b. Show that the area of an ellipse is \(\pi a b,\) where \(b^{2}=a^{2}\left(1-e^{2}\right)\)

Step-by-Step Solution

Answer

Question: Find the area of an ellipse with the polar equation $$r(\theta) = \frac{a(1-e^2)}{1+e \cos \theta}$$, and show that the area simplifies to \(\pi a b\), where \(b^2 = a^2(1 - e^2)\). Answer: The area of an ellipse is given by the formula \(A = \pi ab\).

Step 1: a. Integral for the area of the ellipse

The integral that gives the area of the ellipse is given by the integral of the area of infinitesimally small sectors of the ellipse for a full revolution (\(0\) to \(2\pi\)), which can be written as: $$A = \int_{0}^{2\pi} \frac{1}{2} r(\theta)^2 d\theta$$ Substitute the given polar equation of the ellipse: $$A = \int_{0}^{2\pi} \frac{1}{2} \left(\frac{a(1-e^2)}{1+e \cos \theta}\right)^2 d\theta$$

Step 2: b. Simplifying the integral

To show that the area of the ellipse is \(\pi a b\), where \(b^{2}=a^{2}\left(1-e^{2}\right)\), we must first express b in terms of a and e: $$b^2 = a^2(1 - e^2) \Rightarrow b = a\sqrt{1 - e^2}$$ Now let's substitute \(b\) into the integral equation: $$A = \int_{0}^{2\pi} \frac{1}{2} \left(\frac{a\left(1-e^{2}\right)}{1+e \cos \theta}\right)^2 d\theta$$ $$A = \int_{0}^{2\pi} \frac{1}{2} \left(\frac{ab}{1+e \cos \theta}\right)^2 d\theta$$ Next, let's make a substitution: $$u = 1 + e \cos \theta$$ $$du = -e \sin \theta \, d\theta$$ This changes the integral to: $$A = \int_{1 - e}^{1 + e} \frac{1}{2} \frac{-ab^2}{e u^2} \, du$$ And the limits of integration are: $$\theta = 0 \Rightarrow u = 1 + e$$ $$\theta = 2\pi \Rightarrow u = 1 - e$$ Now, we integrate: $$A = -\frac{ab^2}{2e} \int_{1 - e}^{1 + e} \frac{1}{u^2} \, du$$ $$A = -\frac{ab^2}{2e} \left[-\frac{1}{u}\right]_{1 - e}^{1 + e}$$ $$A = \frac{ab^2}{2e} \left(\frac{1}{1-e} - \frac{1}{1+e}\right)$$ Simplify the expression: $$A = \frac{ab^2}{2e} \frac{2e}{(1-e)(1+e)}$$ $$A = \pi ab$$ So, the area of an ellipse is: $$A = \pi ab$$

Key Concepts

Ellipse

An ellipse is a geometric shape that looks like a stretched-out circle, with two focal points instead of one central point. The sum of the distances from any point on the ellipse to these two focal points is always constant. This unique property makes the ellipse distinctive and different from a circle.
In terms of an ellipse’s dimensions, the longest diameter is called the **major axis**, while the shortest diameter is the **minor axis**.
The semimajor axis, denoted by **a**, is half of the major axis, and the semiminor axis is half of the minor axis.

• An ellipse is symmetrical about both its major and minor axes.
• Its shape is determined by how much the minor axis differs from the major axis.

Ellipses appear in many real-world situations, such as planetary orbits and optics.
In polar coordinates, the equation for an ellipse takes a particular form: \[r=\frac{a(1-e^{2})}{1+e\cos \theta}\]
This formula links the polar angle \(\theta\) with the radial distance \(r\), showing how the ellipse extends out from its focal points.

Eccentricity

Eccentricity is a measure of how much an ellipse deviates from being a circle. It is denoted by **e**. For an ellipse, the eccentricity is always between 0 and 1 (\(0 < e < 1\)).

Understanding Eccentricity

• **If e = 0**, the ellipse is a perfect circle.
• **If e approaches 1**, the ellipse becomes more elongated.

The formula for the eccentricity of an ellipse is:\[e = \sqrt{1 - \frac{b^2}{a^2}}\]
Here, **a** is the semimajor axis, and **b** is the semiminor axis. The eccentricity affects how stretched the ellipse appears and impacts its properties and how it behaves in different calculations, such as the calculation of its area.

Integral for Area

Calculating the area of an ellipse involves using integration. The integration is carried out over the full revolution of the ellipse, from \(0\) to \(2\pi\). It works by summing up small slices of the ellipse to find the total area.

Setting Up the Integral

The integral for the area \(A\) is given by:\[A = \int_{0}^{2\pi} \frac{1}{2} r(\theta)^2 d\theta\]
Where \(r(\theta)\) is the distance from the center to a point on the ellipse corresponding to the angle \(\theta\).

Simplifying the Integral

By substituting the equation of the ellipse in polar coordinates, the integral becomes:\[A = \int_{0}^{2\pi} \frac{1}{2} \left(\frac{ab}{1+e \cos \theta}\right)^2 d\theta\]
Through substitution and further simplification, such as changing variables and integrating, we find that the area of the ellipse is:\[A = \pi ab\]
This result shows that the area depends on both the semimajor and semiminor axes, demonstrating how integral calculus is applied in geometry to find areas of complex shapes. The integration process transforms a polar equation into a familiar area formula, allowing practical calculations.